To find the least positive argument of the fourth root of the complex number \(2 - i\sqrt{12}\), we will follow these steps:
### Step 1: Rewrite the Complex Number
We start with the complex number:
\[
z = 2 - i\sqrt{12}
\]
We can simplify \(\sqrt{12}\) as:
\[
\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}
\]
Thus, we can rewrite \(z\) as:
\[
z = 2 - 2i\sqrt{3}
\]
### Step 2: Convert to Polar Form
Next, we need to convert \(z\) into polar form. The modulus \(r\) of \(z\) is calculated as:
\[
r = |z| = \sqrt{(2)^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4
\]
Now, we find the argument \(\theta\):
\[
\theta = \tan^{-1}\left(\frac{-2\sqrt{3}}{2}\right) = \tan^{-1}(-\sqrt{3})
\]
The angle whose tangent is \(-\sqrt{3}\) is \(-\frac{\pi}{3}\). Since the complex number is in the fourth quadrant, we can express the argument as:
\[
\theta = -\frac{\pi}{3}
\]
### Step 3: Write in Polar Form
Thus, we can express \(z\) in polar form:
\[
z = 4 \left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)
\]
or using Euler's formula:
\[
z = 4 e^{-i\frac{\pi}{3}}
\]
### Step 4: Find the Fourth Root
To find the fourth root of \(z\), we use the formula for the \(n\)-th root of a complex number:
\[
z^{1/4} = r^{1/4} e^{i\frac{\theta + 2k\pi}{n}} \quad (k = 0, 1, 2, 3)
\]
Here, \(r = 4\), \(\theta = -\frac{\pi}{3}\), and \(n = 4\):
\[
z^{1/4} = 4^{1/4} e^{i\frac{-\frac{\pi}{3} + 2k\pi}{4}} = \sqrt{2} e^{i\left(-\frac{\pi}{12} + \frac{k\pi}{2}\right)}
\]
### Step 5: Calculate Arguments for \(k = 0, 1, 2, 3\)
Now we calculate the arguments for \(k = 0, 1, 2, 3\):
- For \(k = 0\):
\[
\text{Argument} = -\frac{\pi}{12}
\]
- For \(k = 1\):
\[
\text{Argument} = -\frac{\pi}{12} + \frac{\pi}{2} = -\frac{\pi}{12} + \frac{6\pi}{12} = \frac{5\pi}{12}
\]
- For \(k = 2\):
\[
\text{Argument} = -\frac{\pi}{12} + \pi = -\frac{\pi}{12} + \frac{12\pi}{12} = \frac{11\pi}{12}
\]
- For \(k = 3\):
\[
\text{Argument} = -\frac{\pi}{12} + \frac{3\pi}{2} = -\frac{\pi}{12} + \frac{18\pi}{12} = \frac{17\pi}{12}
\]
### Step 6: Identify the Least Positive Argument
The least positive argument among these values is:
\[
\frac{5\pi}{12}
\]
### Final Answer
Thus, the least positive argument of the fourth root of the complex number \(2 - i\sqrt{12}\) is:
\[
\boxed{\frac{5\pi}{12}}
\]
